Problem: If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, find the product of those solutions with $a>0.$
Answer: We write $-64 = 2^6 \operatorname{cis} 180^\circ,$ so $x^6 = 2^6 \operatorname{cis} 180^\circ.$  The solutions are of the form
\[x = 2 \operatorname{cis} (30^\circ + 60^\circ k),\]where $0 \le k \le 5.$

[asy]
unitsize(1 cm);

int i;

draw(Circle((0,0),2));
draw((-2.2,0)--(2.2,0));
draw((0,-2.2)--(0,2.2));

dot("$30^\circ$", 2*dir(30), dir(30));
dot("$90^\circ$", 2*dir(90), NE);
dot("$150^\circ$", 2*dir(150), dir(150));
dot("$210^\circ$", 2*dir(210), dir(210));
dot("$270^\circ$", 2*dir(270), SW);
dot("$330^\circ$", 2*dir(330), dir(330));
[/asy]

The solutions where the real part is positive are then $2 \operatorname{cis} 30^\circ$ and $2 \operatorname{cis} 330^\circ,$ and their product is $2 \operatorname{cis} 30^\circ \cdot 2 \operatorname{cis} 330^\circ = 4 \operatorname{cis} 360^\circ = \boxed{4}.$